Calculate the amount of time required to raise temperature of 0.50 liter of water from
0.0° c to 10.0° c if heat energy is supplied at a rate of 1100 joules per second

19 seconds

Heat energy was supplied and used to heat 0.50 liters of water from 0°C to 10°C.

This means water gained heat energy that was supplied.

Therefore;

Heat supplied = Heat gained by water

We are required to calculate the time taken to raise the temperature of water from 0°C to 10°C.

Step 1: Calculate the heat gained by water

Quantity of heat = Mass × specific heat capacity × change in temperature

Density of water 1 g/ml

Volume of water is 500 mL

Therefore, since Mass = density × volume

The mass of water = 500 g

Change in temperature = 10° C

Specific heat capacity of water = 4.18 J/g/°C

Thus;

Quantity of heat = 500 g × 4.18 J/g/°C × 10

                           = 20,900 Joules

Step 2: Heat supplied by a heater

Heat supplied = Power × time

Power = 1100 J/s

Assuming the time required is x

Heat supplied = 1100x Joules

Step 3: Time required

Remember; heat supplied = heat gained by water

Therefore;

1100x joules = 20,900 Joules

      x = 20,900/1100

        = 19 seconds

Therefore, the time required to raise the temperature of water from 0°C to 10°C is 19 seconds.

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answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :